Answer:
(a) The work done by the gas 8.005 x 10⁵ J
(b) the change in internal energy is 5.0575 x 10⁶ J
Explanation:
Given;
heat added to the gas, Q = 1400 kcal = 1400 kcal x 4184 J/kcal = 5.858 x 10⁶ J.
change in volume, ΔV = 19.9 m³ - 12.0 m³ = 7.9 m³
atmospheric pressure, P = 101325 N/m²
(a) The work done by the gas = PΔV
= 101325 x 7.9
= 8.005 x 10⁵ J
(b) the change in internal energy is obtained from first law of thermodynamic;
ΔU = Q - W
ΔU = 5.858 x 10⁶ J - 8.005 x 10⁵ J
ΔU = 58.58 x 10⁵J - 8.005 x 10⁵ J
ΔU = 5.0575 x 10⁶ J
What must the charge (sign and magnitude) of a particle of mass 1.40 gg be for it to remain stationary when placed in a downward-directed electric field of magnitude 640 N/CN/C
Answer:
the charge of the particle is -2.144 x 10⁻⁵ C.
Explanation:
The force acting on the particle is calculated as;
F = EQ = mg
[tex]Q = \frac{mg}{E}[/tex]
where;
Q is magnitude of the charge of the particle
[tex]Q = \frac{(1.4\times 10^{-3})(9.8)}{640} \\\\Q = 2.144 \ \times \ 10^{-5} \ C[/tex]
since the magnetic field is acting downward, the force must be acting upward in opposite direction.
Thus, the charge of the particle will be -2.144 x 10⁻⁵ C.
Complete the following statement: Momentum will be conserved in a two-body collision only if a both bodies come to rest. b the internal forces of the two body system cancel in action-reaction pairs. c the kinetic energy of the system is conserved. d the net external force acting on the two-body system is zero. e the collision is perfectly elastic.
Answer:
d the net external force acting on the two-body system is zero
Explanation:
The net external force acting on the two-body system is zero .
During a collision , two internal forces appear on two bodies in opposite directions separately . As they act on each object separately , they do not cancel each other . But the net force on two body system is zero . Total momentum is always conserved during collision because no external force is involved . Collision is not always elastic .
Please help. I'm stuck!
What is the mass of a catamaran moving at 7.65 m/s that has a momentum of 530145 kg x m/s?
Your family is moving, and you are asked
to help move some boxes. One box is so
heavy that you must push it across the
room rather than lift it. What are some
ways you could reduce friction to make
moving the box easier?
Answer:
Explanation:
I would use a hand truck dollies
What do you think about the attached scenario?
A pion has a rest energy of 135 MeV. It decays into two gamma-ray photons, bursts of electromagnetic radiation that travel at the speed of light. A pion moving through the laboratory at v = 0.98c decays into two gamma-ray photons of equal energies, making equal angles θ with the original direction of motion. Find the angle θ and the energies of the two gamma ray photons.
Answer:
.
Explanation:
.
Energy can be transferred from one place to another through?
What energy store is in the torch
BEFORE it gets switched on?
Answer:
Chemical energy
Explanation:
The energy in the torch is stored as chemical energy before the torch gets switch on.
The chemical energy energy in the battery of cell will power the cell and allows it to produce light.
Chemical energy is a form of potential energy. The electrolytes within the battery are capable of producing electric current. So the chemical energy is transformed into electrical energy which is used to produce the light of the torch.A car pulls on to an onramp with an initial speed of 23.8 mph. The length of the onramp is 852 ft and the car needs to be moving at 45.7 mph at the end of the ramp to merge with traffic. What constant rate of acceleration (in ft/sec2) is required in order to accomplish this
Answer:
The constant rate of acceleration required in order to accomplish this is 1.921 feet per square second.
Explanation:
Let suppose that car accelerates uniformly in a rectilinear motion. Given that initial and final speeds and travelled distances are known, then the acceleration needed by the vehicle ([tex]a[/tex]), measured in feet per square second, is determined by the following kinematic formula:
[tex]a = \frac{v_{f}^{2}-v_{o}^{2}}{2\cdot \Delta x }[/tex] (1)
Where:
[tex]v_{o}[/tex], [tex]v_{f}[/tex] - Initial and final speeds, measured in feet per second.
[tex]\Delta x[/tex] - Travelled distance, measured in feet.
If we know that [tex]v_{o} = 34.907\,\frac{ft}{s}[/tex], [tex]v_{f} = 67.027\,\frac{ft}{s}[/tex] and [tex]\Delta x = 852\,ft[/tex], then acceleration needed to accomplish the task is:
[tex]a = 1.921\,\frac{ft}{s^{2}}[/tex]
The constant rate of acceleration required in order to accomplish this is 1.921 feet per square second.
Scientists believe that the boundary stratum between the Cretaceous and Paleogene was caused by an asteroid. What evidence is most consistent with this theory?
Answer:
Because there was fewer fossils
Explanation:
Answer:
Actually the answer is "The stratum contains iridium.".
Explanation:
Your friend, a world-class long jumper, is trapped on the roof of a burning building. His only escape route is to jump to the roof of the next building. Fortunately for him, he is in telephone contact with you, a Physics 161 student, for advice on how to proceed. He has two options. He can jump to the next building by using the long-jump technique where he jumps at 45o to the horizontal. Or, he can take his chances by staying where he is in the hopes that the fire department will rescue him. You learn from the building engineers that the next building is 10 m away horizontally and the roof is 3 m below the roof of the burning building. You also know that his best long-jump distance is 7.9 m. What do you advise him to do
Answer:
y = -2.69 m
the negative sign indicates that it is descending and the distance is less than the difference in height between the two buildings, therefore the person would be saved in the jump.
Explanation:
his problem must be solved with the missile launch equations.
Let's start by looking for the jumper's initial velocity
R = v₀² sin 2θ / g
for the long jump the angle used is tea = 45º, in the exercise they indicate that the best record is R = 7.9m
v₀² = R g / sin 2te
v₀ = [tex]\sqrt{ \frac{7.9 \ 9.8}{1 }[/tex]
v₀ = 8.80 m / s
Now suppose you jump with this speed to get to the other building, let's use trigonometry for the components of the speed
sin 45 = [tex]v_{oy}[/tex] /v₀
cos 45 = v₀ₓ / v₀
v_{oy} = v₀ sin 45
v₀ₓ = v₀ cos 45
v_{oy} = 8.8 sin 45 = 6.22 m / s
v₀ₓ = 8.8 cos 45 = 6.22 m / s
now let's calculate the sato with these speeds
x = [tex]v_{ox}[/tex] t
the minimum jump is x = 10 m
t = x / v₀ₓ
t = 10 / 6.22
t = 1.61 s
let's find the vertical distance for this time
y = v_{oy} t - ½ g t²
where zero is placed on the jump building
y = 6.22 1.61 - ½ 9.8 1.61²
y = -2.69 m
Let's analyze this result, the negative sign indicates that it is descending and the distance is less than the difference in height between the two buildings, therefore the person would be saved in the jump.
during a typical afternoon thunderstorm in the summer, an area of 66.0 km2 receives 9.57 108 gal of rain in 18 min. how many inches of rain fell during this 18 min period
Answer:
2.16 inch
Explanation:
area under water = 66 km²
= 66 x ( 3280.84 x 12 )² inch²
= 1.023 x 10¹¹ sq inch
volume of rain = 9.57 x 10⁸ gallon = 9.57 x 10⁸ x 231 inch³
= 2.21 x 10¹¹ inch³
If depth of rainfall be t
volume of rain = surface area x depth
= 1.023 x 10¹¹ x t
So ,
1.023 x 10¹¹ x t = 2.21 x 10¹¹
t = 2.16 inch
Select the Moon and use the Info view to determine which of the following statements is correct. The Last Quarter Moon.... rises near noon and sets near midnight. rises at about 6am and sets at about 6pm. rises near midnight and sets near midday. rises and sets at the same time as the Sun. Submit Your Answer
Answer: The correct statement is that the Last Quarter Moon (rises near midnight and sets near midday).
Explanation:
Phases of the moon also called the LUNAR PHASE can be defined as the different shades of illumination on the moon as seen from the earth. The moon is the natural satellite of the earth that illuminates upon reflection of light from the sun. This means it doesn't have power to shine on its own. When carefully observed, there are times it gets dark and beings to glow brighter over a period of time. This occurs because as the moon completes its four weeks lunar cycle round the earth, how much of its face we see illuminated by sunlight depends on the angle the Sun makes with the Moon.
There are 8 main types of the moon phases these includes:
--> New moon: This is when the moon is not visible to the earth because it's between the earth and the sun. It rises at sunrise and sets at sunset.
--> The waxing crescent: At this phase the moon gets brighter and illuminated from the sun that a crescent shape is seen.
--> First quarter: this occurs one week after the new moon. The moon rises at noon and sets at midnight.
--> The waxing gibbous: This occurs after the first quarter phase where more than half of the lit part of the moon is seen.
--> Full moon: This occurs when the moon and the sun are opposite each other. That is way it is said to rise at sunset and sets at sun rise.
--> The waning gibbous: this occurs when more than half of the lit part of the moon gradually becomes darker
--> Third quarter ( Last Quarter): The moon rises at midnight and sets at noon. This occurs a week after the full moon.
--> The waning crescent: This occurs after the last quarter phase where a very thin fading crescent shaped moon is seen, just before the Moon is invisible again at the start of the cycle, the new moon.
A torsional pendulum is formed by attaching a wire to the center of a meter stick with a mass of 5.00 kg. If the resulting period is 4.00 min, what is the torsion constant for the wire
Answer:
The torsion constant for the wire is [tex]2.856\times 10^{-4}\,N\cdot m[/tex].
Explanation:
The angular frequency of the torsional pendulum ([tex]\omega[/tex]), measured in radians per second, is defined by the following expression:
[tex]\omega = \sqrt{\frac{\kappa}{I} }[/tex] (1)
Where:
[tex]\kappa[/tex] - Torsional constant, measured in newton-meters.
[tex]I[/tex] - Moment of inertia, measured in kilogram-square meters.
The angular frequency and the moment of inertia are represented by the following formulas:
[tex]\omega = \frac{2\pi}{T}[/tex] (2)
[tex]I = \frac{m\cdot L^{2}}{12}[/tex] (3)
Where:
[tex]T[/tex] - Period, measured in seconds.
[tex]m[/tex] - Mass of the stick, measured in kilograms.
[tex]L[/tex] - Length of the stick, measured in meters.
By (2) and (3), (1) is now expanded:
[tex]\frac{2\pi}{T} = \sqrt{\frac{12\cdot \kappa}{m\cdot L^{2}} }[/tex]
[tex]\frac{2\pi}{T} = \frac{2}{L}\cdot \sqrt{\frac{3\cdot \kappa}{m} }[/tex]
[tex]\frac{\pi\cdot L}{T} = \sqrt{\frac{3\cdot \kappa}{m} }[/tex]
[tex]\frac{\pi^{2}\cdot L^{2}}{T^{2}} = \frac{3\cdot \kappa}{m}[/tex]
[tex]\kappa = \frac{\pi^{2}\cdot m\cdot L^{2}}{3\cdot T^{2}}[/tex]
If we know that [tex]m = 5\,kg[/tex], [tex]L = 1\,m[/tex] and [tex]T = 240\,s[/tex], then the torsion constant for the wire is:
[tex]\kappa = \frac{\pi^{2}\cdot (5\,kg)\cdot (1\,m)^{2}}{3\cdot (240\,s)^{2}}[/tex]
[tex]\kappa = 2.856\times 10^{-4}\,N\cdot m[/tex]
The torsion constant for the wire is [tex]2.856\times 10^{-4}\,N\cdot m[/tex].
* Psychology
Match the types of psychoactive drugs to their functions,
depressants
stimulants
amphetamines
hallucinogens
to excite neural activity and temporarily
elevate awareness
to increase dopamine activity and produce
schizophrenic-like paranoid symptoms
>
to inhibit the function of the central nervous
system and neural activity
to distort perceptions and effects on thinking
Answer:
See explanation below
Explanation:
Psychoactive drugs are drugs that affect the central nervous system. They alter cognitive function by changing mood and consciousness.
Examples;
Depressants: Inhibit the function of the central nervous system and neural activity.
Stimulants: Excite neural activity and temporarily elevate awareness.
Amphetamines: Increase dopamine activity and produce schizophrenic-like symptoms.
Hallucinogens: Distort perceptions and effects on thinking.
A drug is any substance that alters how the body functions.
What is a drug?A drug is any substance that alters how the body functions. There are different types of drugs that affect different parts of the body.
We shall now explain the following classifications of drugs;
depressants - to inhibit the function of the central nervousstimulants - elevate awarenesshallucinogens - to distort perceptions and effects on thinkingamphetamines - schizophrenic-like paranoid symptomsLearn more about drugs: https://brainly.com/question/6022349
As a laudably skeptical physics student, you want to test Coulomb's law. For this purpose, you set up a measurement in which a proton and an electron are situated 865 nm from each other and you study the forces that the particles exert on each other. As expected, the predictions of Coulomb's law are well confirmed. You find that the forces are attractive and the magnitude of each force is:______
Answer:
force F = 1.66 × [tex]10^{-13}[/tex] N
Explanation:
given data
proton and an electron = 865 nm
solution
we get here force that is express as
force F = k q1 q2 ÷ r² ......................1
put here value and we get
force F = 9 × [tex]10^{9}[/tex] × [tex]\frac{1.6\times (10^{-19})^{2}}{865 \times (10^{-9})^{2}}[/tex]
force F = 1.66 × [tex]10^{-13}[/tex] N
Why does it rain more in West Ferris than in East Ferris? Explain your answer.
Answer:
This idea helps students explain why more rain forms over West Ferris than East Ferris. ... Therefore, when students explain that water vapor condenses higher in the atmosphere, they are actually explaining that water vapor condenses high in the troposphere, which is relatively low in the atmosphere.
Explanation:
Plz mark me brainliest thank u> have a good day
What is average acceleration due to gravity on Earth for a 2000 kg boulder, in proper SI units?
Answer:
9.8m/s²
Explanation:
The average acceleration due to gravity on Earth for a 2000kg boulder is 9.8m/s².
Every object on earth is accelerated towards the center by a rate of change of velocity with time value of 9.8m/s².
The acceleration due to gravity on earth is a constant value from places to places.
For other planetary bodies, the value varies and it differs.
But on earth every object is accelerated at 9.8m/s².
Required
Momentum
The magnitude of the momentum of an object is 64 kg*m/s. If the velocity of the
object is doubled, what will be the magnitude of the momentum of the object? *
32 kg*m/s
64 kg*m/s
128 kg*m/s
256 kg*m/s
Answer:
C) 128 kg*m/s
Explanation:
When you double something you multiply it by 2 most of the time. 64*2=128 or you can add it 64+64=128. Hope this helps.
How do you calculate area when pressure and force are given to you
Answer:
This is my answer
Explanation:
First convert 150 kPa to Pa:
150 × 1,000 = 150,000.
Next substitute the values into the equation:
force normal to a surface area = pressure × area of that surface.
force = 150,000 × 180.
force = 27,000,000 N.
1.First convert 150kPato Pa:
2.150 x 1,000 + 150,000
3.next substitute the values into the equations:
4.force normal to a surface area =pressure x area of that surface.
5.force=150,000 x 180.
6.force = 27,000,000N.
can i have brainliest please
Fluids
A = 2804 cm3 B = 2862 cm2 C = 2916 cm3
Three separate fluids, A, B, and C have been selected at random and each initially fills a 3000 cm3 volume at atmospheric pressure. A gage pressure of 6 x 107 N/m2 is then applied to each fluid. The final volume is given below. Determine which fluids were selected from the given list.
Acetone E = 0.92 GPa Glycerin E = 4.35 GP
Water E = 2.15 GPa Mercury E = 28.5 GPa
Benzene E = 1.05 GPa Sulfuric Acid E = 3.0 GPa
Ethyl Alcohol E = 1.06 GPa Gasoline E = 1.3 GPa
Petrol E = 1.45 GPa Seawater E = 2.34 GPa
Answer:
Explanation:
Fluid A :
Δ V = Change in volume = (3000 - 2804) x 10⁻⁶ m³ = 196 x 10⁻⁶ m³
volume strain = Δ V / V = 196 x 10⁻⁶ / 3000 x 10⁻⁶
= .06533
Δ P = increase in pressure = 6 x 10⁷ Pa
E = Δ P / volume strain = 6 x 10⁷ / .06533 = 91.84 x 10⁷ Pa = .92 GPa .
It is Acetone .
Fluid B :
Δ V = Change in volume = (3000 - 2862) x 10⁻⁶ m³ = 138 x 10⁻⁶ m³
volume strain = Δ V / V = 138 x 10⁻⁶ / 3000 x 10⁻⁶
= .046
Δ P = increase in pressure = 6 x 10⁷ Pa
E = Δ P / volume strain = 6 x 10⁷ / .046 = 130.43 x 10⁷ Pa = 1.3 GPa .
It is Gasoline .
Fluid C :
Δ V = Change in volume = (3000 - 2916) x 10⁻⁶ m³ = 84 x 10⁻⁶ m³
volume strain = Δ V / V = 84 x 10⁻⁶ / 3000 x 10⁻⁶
= .028
Δ P = increase in pressure = 6 x 10⁷ Pa
E = Δ P / volume strain = 6 x 10⁷ / .028 = 214.28 x 10⁷ Pa = 2.14 GPa .
It is Water .
Surface currents are on the
of the Earth's oceans
The radius of the Sun is 6.96 x 108 m and the distance between the Sun and the Earth is roughtly 1.50 x 1011 m. You may assume that the Sun is a perfect sphere and that the irradiance arriving on the Earth is the value for AMO, 1,350 W/m2. Calculate the temperature at the surface of the Sun.
Answer:
5766.7 K
Explanation:
We are given that
Radius of Sun , R=[tex]6.96\times 10^{8} m[/tex]
Distance between the Sun and the Earth, D=[tex]1.50\times 10^{11}m[/tex]
Irradiance arriving on the Earth is the value for AMO=[tex]1350W/m^2[/tex]
We have to find the temperature at the surface of the Sun.
We know that
Temperature ,T=[tex](\frac{K_{sc}D^2}{\sigma R^2})^{\frac{1}{4}}[/tex]
Where [tex]K_{sc}=1350 W/m^2[/tex]
[tex]\sigma=5.67\times 10^{-8}watt/m^2k^4[/tex]
Using the formula
[tex]T=(\frac{1350\times (1.5\times 10^{11})^2}{5.67\times 10^{-8}\times (6.96\times 10^{8})^2})^{\frac{1}{4}}[/tex]
T=5766.7 K
Hence, the temperature at the surface of the sun=5766.7 K
A daredevil is attempting to jump his motorcycle over a line of buses parked end to end by driving up a ramp of 64.8 degrees at a speed of 25.4 m/s. What would be the largest number of buses he can clear if the top of the takeoff ramp is at the same height as the bus tops and the buses are 10.0 m long
Answer: he can only make it over 5 buses
Explanation:
Given the data in the question;
we know that range is expressed as;
R = (V₀²sin2∅₀)/g
V₀ is the initial velocity( 25.4 m/s), ∅₀ is the angle of projection( 64.8°), g is acceleration due to gravity( 9.8 m/s²),
so we substitute
R = ((25.4)²sin2(64.8))/9.8
R = 50.7 m
now, them number of buses will be;
n = R / bus length
given that bus length is 10.0 m
we substitute
n = 50.7 m / 10.0
n = 5.07 ≈ 5
Therefore, he can only make it over 5 buses
Does changing the height of point C affect the speed of the coaster car at point D?
Without friction, NO.
The speed at D depends only on the difference in height between A and D. Whatever happens between them doesn't matter.
The speed of the coaster car at point D will be affected if the height of point C is changed.
Potencial Energy:
It is the enrgy in a body due to the position of differnt part of the object or system.
As we increase the the hight of the car the potetial enrgy increase, the gravitational acceleration on car will be more due to the high of the point C.
Therefore, the speed of the coaster car at point D will be affected if the height of point C is changed.
To know more about speed of the coaster car,
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If the ball that is thrown downward has an acceleration of magnitude aaa at the instant of its release (i.e., when there is no longer any force on the ball due to the woman's hand), what is the relationship between aaa and ggg, the magnitude of the acceleration of gravity
Explanation:
At the instant of release there is no force but an acceleration of a, this means the ball is falling freely under the force of gravity. Then the acceleration would be due to force of gravity and acceleration a = g =9.81 m/s^2.
g= acceleration due to gravity
Which of the following is NOT a step used to perform a scientific inquiry
Answer:
b. Designing an uncontrolled experiment.
Explanation:
They always have it controlled.
Answer:
B. Designing an uncontrolled experiment.
Explanation:
Correct Answer!!!!!!
Under what conditions will a moving 0.030 kg marble and a moving 2.43 kg rock have the same kinetic energy
Answer:
To have the same kinetic energy the speed of the marble must be 9 times the speed of rock.
Explanation:
The general formula of kinetic energy is given as follows:
[tex]K.E = \frac{1}{2}mv^{2}[/tex]
where,
K.E = Kinetic Energy
m = mass of the object
v = speed of the object
So, for the marble and rock to have same kinetic energy, we can write:
[tex]K.E_{marble} = K.E_{rock}\\\\\frac{1}{2}m_{marble}v_{marble}^{2} = \frac{1}{2}m_{rock}v_{rock}^{2}\\\\(0.03\ kg)v_{marble}^{2} = (2.43\ kg)v_{rock}^{2}\\\\taking\ square\ root\ on\ both\ sides:\\v_{marble} = \sqrt{\frac{2.43\ kg}{0.03\ kg}}v_{rock}\\\\v_{marble} = 9\ v_{rock}[/tex]
Hence, to have the same kinetic energy the speed of the marble must be 9 times the speed of rock.
Two 90.0-kilogram people are separated by 3.00 meters. What is the magnitude of the gravitational force that one person exerts on the other?
Answer:
the magnitude of gravitational force is 6 x 10⁻⁸ N.
Explanation:
Given;
mass of the two people, m₁ and m₂ = 90 kg
distance between them, r = 3.0 m
The magnitude of gravitational force exerted by one person on another is calculated as;
[tex]F = \frac{Gm_1m_2}{r^2} \\\\[/tex]
where;
G is gravitational constant = 6.67 x 10⁻¹¹ Nm²/kg²
[tex]F = \frac{Gm_1m_2}{r^2} \\\\F = \frac{6.67\times 10^{-11} \times \ 90 \ \times \ 90}{3^2} \\\\F = 6\times 10^{-8} \ N[/tex]
Therefore, the magnitude of gravitational force is 6 x 10⁻⁸ N.
A large, metallic, spherical shell has no net charge. It is supported on an insulating stand and has a small hole at the top. A small tack with charge Q is lowered on a silk thread through the hole into the interior of the shell.
1) What is the charge on the inner surface of the shell?
A) Q
B) Q/2
C) 0
D) -Q/2
E) -Q
2) What is the charge on the outer surface of the shell?
A) Q
B) Q/2
C) 0
D) -Q/2
E) -Q
3) The tack is now allowed to touch the interior surface of the shell. After this contact, what is the charge on the tack?
A) Q
B) Q/2
C) 0
D) -Q/2
E) -Q
4) What is the charge on the inner surface of the shell now?
A) Q
B) Q/2
C) 0
D) -Q/2
E) -Q
5) What is the charge on the outer surface of the shell now?
A) Q
B) Q/2
C) 0
D) -Q/2
E) -Q
The charge on the inner surface of the shell is -Q
The charge on the outer surface of the shell is Q
After this contact, the charge on the tack is 0
The charge on the inner surface of the shell now is 0
The charge on the outer surface of the shell now is Q
What is the charge on a shell ?The charge on a shell depends on the situation and the conditions of the shell. If the shell is an electrically neutral object, such as a metallic spherical shell, it has no net charge, meaning that the total positive charge is equal to the total negative charge. However, if the shell has an excess or deficit of electrons, it will have a net charge, either positive or negative, depending on whether it has an excess of electrons or a deficit of electrons.
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